A sum of integers

let $x$ be a perfect square and $y$ positive integer less then next perfect squre, so for every positive integer $y$ , $x \leq y < (\sqrt{x} + 1)^2$

$\sqrt x = \left \lfloor \sqrt y \right \rfloor$

than the sum can be calculated as: $(y - x + 1) * \sqrt x$

sum is:

$(4 - 1) * \sqrt 1 + (9 - 4) * \sqrt{4}$
$+ (16 - 9) * \sqrt 9 + (25 - 16) * \sqrt 16$ $+ (36 - 25) * \sqrt 25 + (49 - 36) * \sqrt 36$ $+ (51 - 49) * \sqrt 49 = 217$

Note that for all $x \in \left[k^2,(k + 1)^2\right)$ we have $\left\lfloor \sqrt{x} \right\rfloor = k$ Note that there are $2k + 1$ integers in $\left[k^2,(k + 1)^2\right)$ . So $\begin{aligned} \sum_{k = 1}^{50} \left\lfloor \sqrt{k} \right\rfloor &= \sum_{k = 1}^{6} k(2k + 1) + 2 \cdot 7 \\ &= 2\sum_{k = 1}^{6} k^2 + \sum_{k = 1}^{6} k + 14 \\ &= \frac{6 \cdot 7 \cdot 13}{3} + \frac{6 \cdot 7}{2} + 14 \\ &= \boxed{217} \end{aligned}$

Its simple ,upto a square its a sequence, so we get it as= ,3(1)+5(2)+7(3)+9(4)......+(6x2 +1)(6)+7+7 =,lets generalise for summation up to nth term we get sigma((2n+1)(n)) =2(sigma(n^2))+sigma(n) we get ans as 217

The function ⌊x⌋:R→Z refers to the greatest integer smaller than or equal to x. For example ⌊2.3⌋=2 and ⌊−5⌋=−5. s0 (1)+(1.141)+........... 1(3)+2(5)+3(7)+4(9)+5(11)+6(13)+7(2)=217

All the perfect squares in the series are 1,4,9,16,25,36 and 49. The roots between 1 and 4 will have the value 1 ,between 4 and 9 will have the value 2 and similarly for all the numbers between the squares (From greatest integer function).

We will get the expression

$1+(2 \times 1) + 2 + (4 \times 2) + 3 + (6 \times 3) + 4 + (8 \times 4) + 5 + (10 \times 5) + 6 + (12 \times 6) + 7+7 = 217$

(The last 7 is added because GIF of $\sqrt {50}$ is 7)

Take note for all perfect square numbers, such as follows; 1,4,9,16,25,36,49 By inspection, without using calculator if possible, $\sqrt{2}$ and $\sqrt{3}$ are slightly larger than 1 but slightly less than 2. Thus, the floor value of these are 1. So there are three 1 value.

Let N be the sum of the floor value from $\sqrt{1}$ to $\sqrt50$

N=1(3)+2(5)+3(7)+5(9)+6(11)+7(2) N= 217

The first step is to determine the perfect squares less than 50:

$1,4,9,16,25,36,49$

Now if you take the square root of any number between two perfect squares and use the floor function on it, it will always evaluate to the integer that the less of the perfect roots. In other words, floor $\sqrt{2}$ is 1, floor $\sqrt{3}$ is 1 and then floor $\sqrt{4}$ is 2. Again, using the floor function between $\sqrt{4}$ up to $\sqrt{9}$ will be always be 2. This is due to the fact that, $1<4$ , so the square root of 1 is less than 4 (and any numbers in between). You keep doing this until you have all 50 square roots taken care of - and this all simplifies to:

$3(1) + 5(2) + 7(3) + 9(4) + 11(5) + 13(6) + 2(7) = 217$

Note that the odd numbers are just the distances from one perfect square to the next, excluding the 2 multiplied by 7 (this is because we only go to the square root of 50, and there are only two instances of 7).

Floor will be to nearest square number. The difference between the square numbers increases as 3,5,7 etc (4-1=3, 9-4=5 etc) by which you can work out that the sum is just:

$(1\times 3) +(2\times 5) +(3 \times 7) +(4\times 9) + (5\times 11) + (13\times 6) + (7\times 2)$ $= 217$

3 + $2 \times 5$ +(3 \times 7)+---+(6 \times 13) + (7 \times 2) = 217